Problem 38
Question
Certain sharks can detect an electric field as weak as 1.0 \(\mu\)V\(/\)m. To grasp how weak this field is, if you wanted to produce it between two parallel metal plates by connecting an ordinary 1.5V AA battery across these plates, how far apart would the plates have to be?
Step-by-Step Solution
Verified Answer
The plates need to be 1.5 million meters apart.
1Step 1: Understanding the Electric Field Equation
The electric field (E) created between two parallel plates connected to a battery can be described using the equation \( E = \frac{V}{d} \), where \( V \) is the voltage of the battery, and \( d \) is the distance between the plates. In our case, \( V = 1.5 \, V \) and \( E = 1.0 \, \mu V/m = 1.0 \times 10^{-6} \, V/m \).
2Step 2: Rearranging the Formula for Distance
To find the distance \( d \), we need to rearrange the formula \( E = \frac{V}{d} \). Solving for \( d \), we get \( d = \frac{V}{E} \).
3Step 3: Substitute Known Values
Substitute the known values into the rearranged formula: \(d = \frac{1.5 \, V}{1.0 \times 10^{-6} \, V/m} \).
4Step 4: Calculate the Distance
Perform the calculation: \( d = \frac{1.5}{1.0 \times 10^{-6}} = 1.5 \times 10^6 \, m \). The plates need to be \(1.5 \times 10^6\) meters apart.
Key Concepts
VoltageParallel PlatesDistance CalculationElectric Field Detection
Voltage
Voltage is the measure of electric potential difference between two points. In an electronic circuit, it indicates how much electric potential energy is available to push electrons through the circuit. This potential difference, measured in volts (V), is like the driving force that moves through the circuit.
In the context of parallel plates, the voltage of a battery, such as a 1.5V AA battery, is used. This is what creates an electric field between the plates when connected. Voltage tells us how strong the electrical "push" or "pull" between the plates is.
Understanding voltage helps us arrange components properly to achieve desired electric fields and effects within circuits.
In the context of parallel plates, the voltage of a battery, such as a 1.5V AA battery, is used. This is what creates an electric field between the plates when connected. Voltage tells us how strong the electrical "push" or "pull" between the plates is.
Understanding voltage helps us arrange components properly to achieve desired electric fields and effects within circuits.
Parallel Plates
Parallel plates are two conductive sheets positioned to face each other, usually in a parallel manner, separated by a certain distance. They often serve as a technique for creating stable electric fields.
When connected to a voltage source like a battery, the parallel plates manage to create a uniform electric field in the space between them. This field is constant in strength and direction across the entire area between the plates, which makes them very useful in experiments and various electronics like capacitors.
The field between parallel plates follows the formula: \[ E = \frac{V}{d} \] where \( E \) is the electric field, \( V \) is the voltage applied, and \( d \) signifies the separation distance.
When connected to a voltage source like a battery, the parallel plates manage to create a uniform electric field in the space between them. This field is constant in strength and direction across the entire area between the plates, which makes them very useful in experiments and various electronics like capacitors.
The field between parallel plates follows the formula: \[ E = \frac{V}{d} \] where \( E \) is the electric field, \( V \) is the voltage applied, and \( d \) signifies the separation distance.
Distance Calculation
Calculating the distance between parallel plates can require solving the electric field equation for distance. In our exercise, we want to find out how far apart the plates need to be to create a precise electric field.
To do this, you can rearrange the electric field formula: \[ E = \frac{V}{d} \] to solve for distance \( d \): \[ d = \frac{V}{E} \] Once rearranged, you can substitute in the known values of voltage and the desired electric field strength.
The calculation in the exercise uses a standard AA battery voltage of 1.5V and a target electric field of \(1.0 \times 10^{-6} \, V/m\), resulting in a very large distance (\(1.5 \times 10^6\) meters) to achieve such a weak field.
To do this, you can rearrange the electric field formula: \[ E = \frac{V}{d} \] to solve for distance \( d \): \[ d = \frac{V}{E} \] Once rearranged, you can substitute in the known values of voltage and the desired electric field strength.
The calculation in the exercise uses a standard AA battery voltage of 1.5V and a target electric field of \(1.0 \times 10^{-6} \, V/m\), resulting in a very large distance (\(1.5 \times 10^6\) meters) to achieve such a weak field.
Electric Field Detection
Electric field detection involves sensing the presence and strength of an electric field in an area. Certain animals, like sharks, have the natural ability to detect these fields, even when they are extremely weak.
In physics and engineering, being able to detect an electric field allows for understanding how charges and voltages distribute around various devices.
The field is detected by its effect on charged particles or electronic sensors. The amount of force experienced by a charged particle in a field reveals the strength of that field. This detection is critical in measuring and understanding electric phenomena in diverse applications such as sensors and electrostatic experiments.
In physics and engineering, being able to detect an electric field allows for understanding how charges and voltages distribute around various devices.
The field is detected by its effect on charged particles or electronic sensors. The amount of force experienced by a charged particle in a field reveals the strength of that field. This detection is critical in measuring and understanding electric phenomena in diverse applications such as sensors and electrostatic experiments.
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